Is this dynamic array allocation allowed

Sarfaraz Nawaz Over a year ago

I believe C++98 did not support value initialized operator new[]

You can write new int[n[0]]() in C++11 as well. :P

Sarfaraz Nawaz Over a year ago

Also, none of the syntaxes uses "default" initialization. They're "value" initializations. And there is a huge difference.

Cornstalks Over a year ago

@Nawaz: I know, I wasn't trying to say you couldn't do that. What I was trying to show is syntax that different versions of the C++ standard introduced. C++98 introduced (), C++11 introduced {}. And I know they're value initializations. That's why I said "value" first in "value/default initializations". Many of the people I've talked to say "default initialize" though, so I included that term as it's more easily understood by many of the people I've talked to. A misnomer, sure, but colloquial none the less.

Sarfaraz Nawaz Over a year ago

All I meant, if you're writing "value/default" initialization, you're being incorrect, or at least imprecise and confusing.

|

Yu Hao · Accepted Answer · 2014-10-14 15:36:30Z

3

Yes, it's valid.

It's uninitalized. To initialize the elements to 0, use

 int *array1 = new int[n[0]]();
 //                         ^^

answered Oct 14, 2014 at 15:36

Yu Hao

123k50 gold badges252 silver badges305 bronze badges

Comments

Paul R · Accepted Answer · 2014-10-14 15:41:24Z

3

Yes, it's fine, but if you want the array data to be initialised to zero then you need:

int *array1 = new int[n[0]]();
int *array2 = new int[n[1]]();
                          ^^^^

You might want to consider using slightly more up-to-date C++ idioms though - std::vector would probably be a better choice than raw arrays:

std::vector<int> array1(n[0]);
std::vector<int> array2(n[1]);

edited Oct 14, 2014 at 15:41

answered Oct 14, 2014 at 15:35

Paul R

214k38 gold badges402 silver badges579 bronze badges

8 Comments

Thanks to all who answered. The added parenthesis you have at the end of the new statement initializes elements to 0?

Paul R Over a year ago

Yes, that's default initialisation syntax.

Thats odd, I still print out large obscure numbers even with those parenthesis ..

drescherjm Over a year ago

Are you going outside the bounds of the array?

Paul R Over a year ago

I think I'd need to see the current version of the code in order to understand why this isn't working for you.

|

Sergey Kalinichenko · Accepted Answer · 2014-10-14 15:42:05Z

3

Is this valid for creating array1 size 1000 and array2 size 5000?

Yes, it is valid, because you are using operator new[]. If you declared int array1[n[0]], you would be relying on an extension, because variable-length arrays are not supported in C++.

Will all array positions be set to NULL or 0?

Neither. The values will be undefined (that's a fancy way of saying "garbage") until you assign to them. You can force initialization by appending a pair of parentheses after the call of new[].

edited Oct 14, 2014 at 15:42

answered Oct 14, 2014 at 15:37

Sergey Kalinichenko

729k85 gold badges1.2k silver badges1.6k bronze badges

2 Comments

Sergey Kalinichenko Over a year ago

Variable-length arrays aren't valid in C++14 either (although they were proposed at one point).

@MikeSeymour Thanks! That's nice to know. I thought we'd finally get in C++ what's been available in C for 15 years, but I guess we would have to wait :-(

Slava · Accepted Answer · 2014-10-14 15:41:39Z

2

Yes operator new[] is doing dynamic memory allocation and it is valid to pass a value whatever way you get it. No value will not be initialized and you should not forget to call delete[] for that pointer or use a smart pointer.

Better way would be to use std::vector -

std::vector<int> array1( n[0] );
std::vector<int> array2( n[1] );

you do not need to worry on memory deallocation and data will be initialized. If you need to initialize some different value than 0, just pass it as a second parameter:

std::vector<int> array1( n[0], 123 );
std::vector<int> array2( n[1], 456 );

answered Oct 14, 2014 at 15:41

Slava

44.4k2 gold badges54 silver badges100 bronze badges

6 Comments

If I maintain variable names array1, wouldn't it be better to use arrays instead of vectors? Because after freeing, I can store whatever I want into that array again. But with vectors, there is no free statement?

@user3507072: No, you can't use an array after freeing it. It no longer exists.

Correct, but another declaration and initialization of the array solves this. Are you implying that vectors automatically free the old version when re-declared?

@user3507072: I've no idea what you mean. You can't declare the pointer a second time. You could reassign it to point to a newly allocated array, just as you could reassign (or resize) a vector.

Baldrickk Over a year ago

@user3507072 to start with free() goes with malloc() not new (you're looking for delete[]). With vectors, all the memory management is done for you; vector goes out of scope, and destructors are called automatically. You can change the data in a vector as easilly as in an array but you can't use either once deleted or out of scope, and finally, you can call the variables whatever you want, std::vector<int> supercalifragilisticexpealidocious if you wish. Call it whatever is suitable for the problem.

|

Community · Accepted Answer · 2017-05-23 12:05:50Z

1

Is this valid for creating array1 size 1000 and array2 size 5000?

Yes. But you'll be better off using automatically managed dynamic arrays:

std::vector<int> array1(1000);

to save yourself the bother of remembering to delete the array once you've finished with it. It can be surprisingly difficult to get that right without RAII.

Will all array positions be set to NULL or 0?

No. new will default-initialise the objects it creates; in the case of primitive types like int, default-initialisation doesn't do anything, leaving them with indeterminate values.

If you want them to be value-initialised to zero, then you can specify that:

int *array1 = new int[n[0]]();
                           ^^

or use vector, which will value-initialise them unless you specify another initial value.

edited May 23, 2017 at 12:05

CommunityBot

11 silver badge

answered Oct 14, 2014 at 15:40

Mike Seymour

256k30 gold badges467 silver badges658 bronze badges

5 Comments

value-initialized is not standard for c++98, is it?

@Slava: Maybe not; I can't remember that far back.

I believe it is supported by gcc but it is a non standard extension for c++98

@Slava: I don't see the relevance of historical trivia to this question. Value-initialisation has been standard for more than a decade; in any case, you should be using vector rather than juggling pointers.